Question

for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals where the function is increasing and decreasing, x - coordinates of the inflection points, open intervals where the function is concave up and concave down, and relative minima and maxima. using this information, sketch the graph of the function.

1. ( y = -\frac{x^{3}}{3}+x^{2} )

Explanation:

Step1: Find x-intercepts

Set $y=0$: $-\frac{x^3}{3} + x^2 = 0 \Rightarrow x^2(-\frac{x}{3} + 1) = 0 \Rightarrow x=0$ or $x=3$

Step2: Find y-intercept

Set $x=0$: $y=0$

Step3: Compute first derivative

$y' = -x^2 + 2x$

Step4: Find critical points

Set $y'=0$: $-x^2 + 2x = 0 \Rightarrow x(-x + 2)=0 \Rightarrow x=0$ or $x=2$

Step5: Determine increasing/decreasing intervals

Test $y'$:

• $x<0$: $y'<0$ (decreasing)
• $00$ (increasing)
• $x>2$: $y'<0$ (decreasing)

Step6: Find relative extrema

• $x=0$: $y'$ changes from - to + → relative minimum
• $x=2$: $y'$ changes from + to - → relative maximum

Compute values: $y(0)=0$, $y(2)=-\frac{8}{3}+4=\frac{4}{3}$

Step7: Compute second derivative

$y'' = -2x + 2$

Step8: Find inflection points

Set $y''=0$: $-2x + 2=0 \Rightarrow x=1$
$y(1)=-\frac{1}{3}+1=\frac{2}{3}$

Step9: Determine concavity intervals

Test $y''$:

• $x<1$: $y''>0$ (concave up)
• $x>1$: $y''<0$ (concave down)

Answer:

• x-intercepts: $x=0, 3$
• y-intercept: $y=0$
• Critical points: $x=0, 2$
• Increasing interval: $(0,2)$
• Decreasing intervals: $(-\infty,0)\cup(2,\infty)$
• Relative minimum: $(0,0)$
• Relative maximum: $(2,\frac{4}{3})$
• Inflection point: $(1,\frac{2}{3})$
• Concave up: $(-\infty,1)$
• Concave down: $(1,\infty)$

(Graph: Plot intercepts (0,0),(3,0), extrema (0,0),(2,4/3), inflection point (1,2/3); curve decreases from left to (0,0), increases to (2,4/3), decreases after; concave up before x=1, concave down after.)